Calculus II Test #1 March 5, 2002 Name____________________ R.  Hammack Score ______

(1) Evaluate the following integrals.
(a)

(b) 2 22tan(u) +C = 2 tan()+C

(c) -2/3 +C = - +C

u = cos x
du = -sin x dx

(d) 4/2 - 1/2 = 3/2

u = ln x
du = 1/x dx

(2) A object moving on a straight line has a (nonconstant) acceleration of  6t  feet per second per second at time t. It travels a distance of 33 feet between times t = 0 and t = 3.  Find the object's velocity function v(t).
a(t) = 6t
v(t) =
= 3+C        (We just need to find C)
s(t) = = +Ct +K
From s(0) = 0 we get K = 0, so  s(t) = +Ct Now, 33 = s(3) = gives the equation 33 = 27 + 3C, or  3C = 6, so C = 2.
Thus v(t) =  3+2

(3)

(4) Find the average value of the function on the interval [0, ln(2)].
Using the formula for average value, the average value is
= ==

(5) 20
Over the interval [6,2], the graph of y = |x| forms two triangles. Adding their areas, we get the answer of
(1/2)(2)(2) + (1/2)(6)(6) = 2 + 18 = 20.

(6)  The graph of a function f(x) has slope   at each point (x, f(x)).
Also, the graph of  f(x) passes through the point (1, 5).  Find the function f(x).

The information says slope = f '(x) = .
This means  f(x) =
.
Thus f(x) =
To determine f completely, we just need to find C. To do this, use the fact that f(1)=5.
5=f(1)=
5= 0 + 3/4 +C
Therefore C = 17/4, and  the answer to the question is f(x) =

(7)  The expression      represents a definite integral over the interval [3, 5].   Write the definite integral. (You do not need to find its value.)    ANSWER: